The electric field is thus seen to depend only on the charge
$Q$ and the distance
$r$ ; it is completely independent of the test charge
$q$ .
Calculating the electric field of a point charge
Calculate the strength and direction of the electric field
$E$ due to a point charge of 2.00 nC (nano-Coulombs) at a distance of 5.00 mm from the charge.
Strategy
We can find the electric field created by a point charge by using the equation
$E=\text{kQ}/{r}^{2}$ .
Solution
Here
$Q=2\text{.}\text{00}\times {\text{10}}^{-9}$ C and
$r=5\text{.}\text{00}\times {\text{10}}^{-3}$ m. Entering those values into the above equation gives
This
electric field strength is the same at any point 5.00 mm away from the charge
$Q$ that creates the field. It is positive, meaning that it has a direction pointing away from the charge
$Q$ .
Calculating the force exerted on a point charge by an electric field
What force does the electric field found in the previous example exert on a point charge of
$\mathrm{\u20130.250}\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ ?
Strategy
Since we know the electric field strength and the charge in the field, the force on that charge can be calculated using the definition of electric field
$\mathbf{\text{E}}=\mathbf{\text{F}}/q$ rearranged to
$\mathbf{\text{F}}=q\mathbf{\text{E}}$ .
Solution
The magnitude of the force on a charge
$q=-0\text{.}\text{250}\phantom{\rule{0.25em}{0ex}}\text{\mu C}$ exerted by a field of strength
$E=7\text{.}\text{20}\times {\text{10}}^{5}$ N/C is thus,
Because
$q$ is negative, the force is directed opposite to the direction of the field.
Discussion
The force is attractive, as expected for unlike charges. (The field was created by a positive charge and here acts on a negative charge.) The charges in this example are typical of common static electricity, and the modest attractive force obtained is similar to forces experienced in static cling and similar situations.
Play ball! Add charges to the Field of Dreams and see how they react to the electric field. Turn on a background electric field and adjust the direction and magnitude.
Section summary
The electrostatic force field surrounding a charged object extends out into space in all directions.
The electrostatic force exerted by a point charge on a test charge at a distance
$r$ depends on the charge of both charges, as well as the distance between the two.
The electric field
$\mathbf{\text{E}}$ is defined to be
$\mathbf{\text{E}}=\frac{\mathbf{\text{F}}}{q,}$
where
$\mathbf{\text{F}}$ is the Coulomb or electrostatic force exerted on a small positive test charge
$q$ .
$\mathbf{\text{E}}$ has units of N/C.
The magnitude of the electric field
$\mathbf{\text{E}}$ created by a point charge
$Q$ is
What is the magnitude and direction of an electric field that exerts a
$2\text{.}\text{00}\times {\text{10}}^{-5}\phantom{\rule{0.25em}{0ex}}\text{N}$ upward force on a
$\mathrm{\u20131.75}\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ charge?
What is the magnitude and direction of the force exerted on a
$3.50\phantom{\rule{0.25em}{0ex}}\mu \text{C}$ charge by a 250 N/C electric field that points due east?
Calculate the initial (from rest) acceleration of a proton in a
$5\text{.}\text{00}\times {\text{10}}^{6}\phantom{\rule{0.25em}{0ex}}\text{N/C}$ electric field (such as created by a research Van de Graaff). Explicitly show how you follow the steps in the Problem-Solving Strategy for electrostatics.
(a) Find the direction and magnitude of an electric field that exerts a
$4\text{.}\text{80}\times {\text{10}}^{-\text{17}}\phantom{\rule{0.25em}{0ex}}\text{N}$ westward force on an electron. (b) What magnitude and direction force does this field exert on a proton?
Since you said they have the same momentum.. So meaning that there is more like an inverse proportionality in the quantities used to find the momentum. We are told that the the is a larger mass and a smaller mass., so we can conclude that the smaller mass had higher velocity as compared to other one
Gift
Mathamaticaly correct
megavado
Mathmaticaly correct :)
megavado
I have proven it by using my own values
Gift
Larger mass=4g
Smaller mass=2g
Momentum of both=8
Meaning V for L =2 and V for S=4
Now find there kinetic energies using the data presented
Gift
grateful soul...thanks alot
Faith
Welcome
Gift
2 stones are thrown vertically upward from the ground, one with 3 times the
initial speed of the other. If the faster stone takes 10 s to return to the ground, how
long will it take the slower stone to return? If the slower stone reaches a maximum
height of H, how high will the faster stone go
Suppose that a grandfather clock is running slowly; that is, the time it takes to complete each cycle is longer than it should be. Should you (@) shorten or (b) lengthen the pendulam to make the clock keep attain the preferred time?
shorten it, since that is practice able using the simple pendulum as experiment
Silvia
it'll always give the results needed no need to adjust the length, it is always measured by the starting time and ending time by the clock
Paul
it's not in relation to other clocks
Paul
wat is d formular for newton's third principle
Silvia
okay
Silvia
shorten the pendulum string because the difference in length affects the time of oscillation.if short , the time taken will be adjusted.but if long ,the time taken will be twice the previous cycle.
Boyles law states that the volume of a fixed amount of gas is inversely proportional to pressure acting on that given gas if the temperature remains constant
which is:
V<k/p or
V=k(1/p)
FADILAT
how to convert meter per second to kilometers per hour